Question

Which transition in the hydrogen atomic spectrum will have the same wavelength as the Balmer transition (i.e. $n=4$ to $n=2$ ) of $H{e}^{+}$spectrum?

• A. n=4 to n=3
• B. n=3 to n=2
• C. n=4 to n=2
• D. n=2 to n=1

Answer 1

Answer: (D)

$\because \frac{1}{\lambda }=Z^2\cdot R\left[\frac{1}{n_1^2}-\frac{1}{n_2^2}\right]$
$\therefore$ For $H{e}^{+}$ion, $Z=2$
$\frac{1}{\lambda }=2^2\cdot R\left[\frac{1}{2^2}-\frac{1}{4^2}\right]$
$\frac{1}{\lambda }=4\times R\times \frac{3}{16}=\frac{3}{4}R$
The same value for $H$-atom is possible when electron jumps from $n=2$ to $n=1$, i.e.
$\frac{1}{\lambda }=1\times R\left[\frac{1}{1}-\frac{1}{4}\right]$
$\Rightarrow \frac{3}{4}R$